Could someone please help me with this math problem

Sammy hita a massive pop-up. the highth,in metres of the ball is given by the equation
h=-5t^2 +20t +1.2, where t is in seconds.
a)find the maximum hight of the ball? solve algebrically
b)at what hight did sammy hit the ball
c) if the ball is caught at the same hight at which it was hit, how long is it in the air
d)if the ball hits the ground, how long is it in the air? solve algebrically

i really don't understand how to do it and help will be very appreciated

It might be instructive to plot the curve, plug in some values of t and get out some values of h.
Draw t as the horizontal axis and h as the vertical axis.
What you will find is that the equation describes a parabola trajectory. The maximum of this trajectory can be found by differentiating h with respect to t to give:

dh/dt = -10t +20

equating this to 0 (the gradient at the maximum) gives:

-10t + 20 = 0 or t = 2 seconds at the maximum.

Plug t = 2 back in to your equation for h to give the maximum height as:

h = -5(2)^2 +20(2) +1.2 = 21.2 metres. a)

Sammy hit the ball a t = 0 so plug that in to give h = 1.2 metres b)

If it was caught at this height then solve for the case where the quadratic intersect the horizontal line h = 1.2 or:

5t^2 + 20t = 0 roots are t1= 0 as expected and t2 = 20/5 = 4 seconds c)

If it hits the ground find the other roots of your quadratic when equating to 0:

-5t^2 + 20t +1.2 = 0 roots are t1 = -0.059 seconds (ignore) and t2 = 4.059 seconds d)

a) write the equation in vertex form by completing the square. the y-value of the vertex is the maximum height of the ball
b) the height that the ball was hit from will be the only term in the equation that does not change with respect to time (i.e. the constant)
c) set h equal to the constant term you found in part (b) and solve for t (quadratic formula, probably)